Problem 5
Question
Determine whether or not the given equations are quadratic. If the resulting form is quadratic, identify \(a, b,\) and \(c,\) with \(a>0 .\) Otherwise, explain why the resulting form is not quadratic. $$x^{2}=(x+2)^{2}$$
Step-by-Step Solution
Verified Answer
The equation simplifies to a linear form, so it is not quadratic.
1Step 1: Expand the Equation
Start by expanding the right side of the equation \((x + 2)^2\) using the square of a binomial formula:\[(x+2)^2 = x^2 + 4x + 4\].Thus, the original equation becomes:\[x^2 = x^2 + 4x + 4\].
2Step 2: Simplify the Equation
Subtract \(x^2\) from both sides of the equation to simplify it:\[x^2 - x^2 = x^2 + 4x + 4 - x^2\],which simplifies to:\[0 = 4x + 4\].
3Step 3: Determine the Form of the Equation
Now the equation is \(0 = 4x + 4\), which is a linear equation, not a quadratic equation. A quadratic equation has the form \(ax^2 + bx + c = 0\) with \(a eq 0\). Since there is no squared term left, it cannot be quadratic.
Key Concepts
Quadratic EquationBinomial ExpansionEquation Simplification
Quadratic Equation
A quadratic equation is a type of polynomial equation with the highest degree term being a square (\(x^2\)) term. It generally takes the form:\[ax^2 + bx + c = 0\]where \(a\), \(b\), and \(c\) are constants, with \(a eq 0\). This ensures that the equation "curves" in a parabolic shape. Quadratics are significant in algebra as they often appear in various applications, including physics, engineering, and even finance.
- **Standard Form:** The standard form of a quadratic is \(ax^2 + bx + c = 0\).
- **Coefficients:** The values \(a, b, c\) are the equation's coefficients. \(a\) is the leading coefficient and must be non-zero.
- **Solution Methods:** Quadratics can be solved using various methods such as factoring, completing the square, or the quadratic formula.
Binomial Expansion
The expansion of a binomial expression like \((x + 2)^2\) involves distributing and rearranging terms. Here, binomial expansion refers to transforming an expression with two terms raised to power into a polynomial. In the case of \((x + 2)^2\), this can be expanded using the square of a binomial formula:\[(x + 2)^2 = x^2 + 2 \cdot 2x + 2^2\]Breaking this down yields:
- **First Term:** \(x^2\), from multiplying \(x\) by itself.
- **Second Term:** \(2x\) twice, which results in \(4x\), due to cross-multiplication of terms (\(x\) and \(2\)).
- **Third Term:** \(4\), which comes from \(2 \times 2\).
Equation Simplification
In algebra, simplifying equations is crucial to solving them efficiently and correctly. This involves manipulating the equation, often by combining like terms, and performing valid operations to make it more manageable. In the given problem, simplifying began by aligning terms from both sides after expanding \((x + 2)^2\):\[x^2 = x^2 + 4x + 4\]The goal was to simplify the terms by subtraction. Notice how subtracting \(x^2\) from each side simplifies to:\[0 = 4x + 4\]At this point, the equation transforms from a quadratic potential to a linear one.
- **Identifying Forms:** Quadratic versus linear depends on the presence of the \(x^2\) term.
- **Simplification Goals:** Removing terms that cancel each other can change the equation type.
- **Equation Forming:** This simplification clarified that the equation was fundamentally linear, as no squared terms remain.
Other exercises in this chapter
Problem 5
Solve the given quadratic equations using the quadratic formula. If there are no real roots, state this as the answer. Exercises \(3-6\) are the same as Exercis
View solution Problem 5
Solve the given quadratic equations by using the square root property. $$x^{2}=7$$
View solution Problem 6
Solve the given quadratic equations using the quadratic formula. If there are no real roots, state this as the answer. Exercises \(3-6\) are the same as Exercis
View solution Problem 6
Sketch the graph of each parabola by using only the vertex and the \(y\) -intercept. Check the graph using a calculator. $$s=2 t^{2}+8 t-5$$
View solution